The voltage in a battery is 9 volts. The resistance is 3 ohms. What is the current?

There are a couple of ways to work this one. You can find the so-called branch currents (the current through each resistor) and then add them to get total current, or you can find the total effective resistance and then apply that information to the applied voltage to compute total current. Let's do this. I = E/R for each resistor. IR1 = E/6 = 9/6 = 3/2 = 1 1/2 amps IR2 = E/3 = 9/3 = 3 amps Itotal = IR1 + IR2 = (1 1/2) + (3) = 4 1/2 amps That's one way. Here's the other. Rtotal = 1 / [(1/R1) + (1/R2)] = 1 / [(1/6) + (1/3)] = 1 / [(1/6) + (2/6)] = 1 / [3/6] = 2 ohms Itotal = Eapplied / Rtotal = 9 volts / 2 ohms = 4 1/2 amps It checks. What a surprise. Ohm's law is correct.